Question: Solve the equation. $\dfrac{dy}{dx}=2xy^{-2}$ Choose 1 answer: Choose 1 answer: (Choice A) A $y=\sqrt[3]{3x^2+C}$ (Choice B) B $y=C\sqrt[3]{3x^2}$ (Choice C) C $y=\sqrt{2x+C}$ (Choice D) D $y=C\sqrt{2x}$
Answer: We can bring this equation to the form $f(y)\,dy=g(x)\,dx$ : $\begin{aligned} \dfrac{dy}{dx}&=2xy^{-2} \\\\ \dfrac{dy}{dx}&=\dfrac{2x}{y^2} \\\\ y^2\,dy&=2x\,dx \end{aligned}$ This means we can solve this equation using separation of variables! $\begin{aligned} y^2\,dy&=2x\,dx \\\\ \int y^2\,dy&=\int 2x\,dx \\\\ \dfrac{y^3}{3}&=x^2+C_1 \\\\ y^3&=3x^2+C \\\\ y&=\sqrt[3]{3x^2+C} \end{aligned}$ [Where did we get C?] Notice that after the integration, more work was required in order to isolate $y$. In conclusion, this is the solution of the equation: $y=\sqrt[3]{3x^2+C}$